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Just from a coordinate standpoint, without regard to application.. if the $x-$ and $x’-$axes are coincident and the other axes are parallel, then $y=y’$, $z=z’$. It’s true that the infinite $xy-$ and $x’y’-$planes, as well as $xz$ and $x’z’$ would cover the same points, but the points would have different values for X coordinate, and not be the same points. The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a+k, b, c)$ in $X’Y’Z’$, where before movement $k$ is constant.

So it is generally true even before looking at the application that $y=y’$.

For the initial set-up, just take a single $x$ axis, and that passes (perpendicularly of course) through the $yz$-plane at a differently numbered point than on the $y’z’$ plane. Those two planes are otherwise coincident, ie are coincident. The $xy$ and $x’y’$ planes are parallel, as are $xz$ and $x’z’$.

If it is moving, we allow $j,k$ to change. Even before any physics, we know they are not moving relatively in the $x$ direction. From a special relativity standpoint this means that the coordinate systems could be moving relatively on $y$ and/or $z$. And $v=\sqrt{\frac{d y}{dt}^2 +\frac{d z}{dt}^2}$ according to both of them. They will not measure objects as having the same length in those directions and will have time dilation if one is faster from a still frame.

Just from a coordinate standpoint, without regard to application.. if the $x-$ and $x’-$axes are coincident and the other axes are parallel, then $y=y’$, $z=z’$. It’s true that the infinite $xy-$ and $x’y’-$planes, as well as $xz$ and $x’z’$, would cover the same points, but the points would have different values for their X coordinate. The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a+k, b, c)$ in $X’Y’Z’$, where before movement $k$ is constant.

So yes it is generally true even before looking at the application that $y=y’$.

If it is moving, we allow $k$ to change. Even before any physics, we know they are not moving relatively in the $Y$ or $Z$ directions. From a special relativity standpoint this means that the coordinate systems would be moving relatively $v=\frac{d k}{dt}$ according to both of them. 

They will not measure objects as having the same length in the $X$ direction and will have time dilation if one is faster from a still frame.

But length contraction is directional, and parallel to the relative motion. The planes will remain parallel. We know this because to change that fact, two points in the (wlog) $XY$ plane would have to move different amounts in the $Z$ direction. This cannot happen as there is no length contraction in that direction.

Just from a coordinate standpoint, without regard to application.. if the $x$- and $x’$-axes are coincident and the other axes are parallel, then $y=y’$, $z=z’$. It’s true that the infinite planes $x=0$ and $x’=0$ would cover the same points, but the points would have different values for the other two coordinates, and not be the same points (ie different coordinates). The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a, b+j, c+k)$ in $X’Y’Z’$, where $j,k$ are constants.

So it’s not generally true even before looking at the application that $y=y’$.

For the initial set-up, just take a single $x$ axis, and that passes (perpendicularly of course) through the $yz$-plane at a differently numbered point than on the $y’z’$ plane. Those two planes are otherwise coincident, ie are coincident. The $xy$ and $x’y’$ planes are parallel, as are $xz$ and $x’z’$.

If it is moving, we allow $j,k$ to change. Even before any physics, we know they are not moving relatively in the $x$ direction. From a special relativity standpoint this means that the coordinate systems could be moving relatively on $y$ and/or $z$. And $v=\sqrt{\frac{d y}{dt}^2 +\frac{d z}{dt}^2}$ according to both of them. They will not measure objects as having the same length in those directions and will have time dilation if one is faster from a still frame.

Just from a coordinate standpoint, without regard to application.. if the $x-$ and $x’-$axes are coincident and the other axes are parallel, then $y=y’$, $z=z’$. It’s true that the infinite $xy-$ and $x’y’-$planes, as well as $xz$ and $x’z’$ would cover the same points, but the points would have different values for X coordinate, and not be the same points. The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a+k, b, c)$ in $X’Y’Z’$, where before movement $k$ is constant.

So it is generally true even before looking at the application that $y=y’$.

For the initial set-up, just take a single $x$ axis, and that passes (perpendicularly of course) through the $yz$-plane at a differently numbered point than on the $y’z’$ plane. Those two planes are otherwise coincident, ie are coincident. The $xy$ and $x’y’$ planes are parallel, as are $xz$ and $x’z’$.

If it is moving, we allow $j,k$ to change. Even before any physics, we know they are not moving relatively in the $x$ direction. From a special relativity standpoint this means that the coordinate systems could be moving relatively on $y$ and/or $z$. And $v=\sqrt{\frac{d y}{dt}^2 +\frac{d z}{dt}^2}$ according to both of them. They will not measure objects as having the same length in those directions and will have time dilation if one is faster from a still frame.

Just from a coordinate standpoint, without regard to application.. if the $x$- and $x’$-axes are coincident, then $x=x’$. It’s true that the infinite planes $x=0$ and $x’=0$ would cover the same points, but the points would have different values for the other two coordinates, and not be the same points (ie different coordinates). The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a, b+j, c+k)$ in $X’Y’Z’$, where $j,k$ are constants.

So it’s not generally true even before looking at the application that $y=y’$.

For the initial set-up, just take a single $x$ axis, and that passes (perpendicularly of course) through the $yz$-plane at a differently numbered point than on the $y’z’$ plane. The $xy$ and $x’y’$ planes are parallel, as are $xz$ and $x’z’$.

If it is moving, we allow $j,k$ to change. Even before any physics, we know they are not moving relatively in the $x$ direction. From a special relativity standpoint this means that the coordinate systems could be moving relatively on $y$ and/or $z$. And $v=\sqrt{\frac{d y}{dt}^2 +\frac{d z}{dt}^2}$ according to both of them. They will not measure objects as having the same length in those directions and will have time dilation if one is faster from a still frame.

Just from a coordinate standpoint, without regard to application.. if the $x$- and $x’$-axes are coincident and the other axes are parallel, then $y=y’$, $z=z’$. It’s true that the infinite planes $x=0$ and $x’=0$ would cover the same points, but the points would have different values for the other two coordinates, and not be the same points (ie different coordinates). The set-up is: any point $(a, b, c)$ in $XYZ$ would equal $(a, b+j, c+k)$ in $X’Y’Z’$, where $j,k$ are constants.

So it’s not generally true even before looking at the application that $y=y’$.

For the initial set-up, just take a single $x$ axis, and that passes (perpendicularly of course) through the $yz$-plane at a differently numbered point than on the $y’z’$ plane. Those two planes are otherwise coincident, ie are coincident. The $xy$ and $x’y’$ planes are parallel, as are $xz$ and $x’z’$.

If it is moving, we allow $j,k$ to change. Even before any physics, we know they are not moving relatively in the $x$ direction. From a special relativity standpoint this means that the coordinate systems could be moving relatively on $y$ and/or $z$. And $v=\sqrt{\frac{d y}{dt}^2 +\frac{d z}{dt}^2}$ according to both of them. They will not measure objects as having the same length in those directions and will have time dilation if one is faster from a still frame.